I have come across many questions where I'm asked to give the number of possible structural isomers. For example number, structural isomers of hexane is 5, while the number structural isomers of decane is 75.

How can I determine the possible number of structural isomers of a given organic compound?

$\begingroup$Is this for a specific class of organic molecule or any organic molecule? Because the numbers you gave suggest that the number of isomers quickly increases with size of molecule, and that's just for simple hydrocarbons. And do stereoisomers count? I tried to come up with an equation for simple hydrocarbons, but couldn't identify 5 isomers for hexane, it's too late at night.$\endgroup$
– user137Sep 10 '14 at 5:32

$\begingroup$@ashu This only gives you the degree of unsaturation, which is for alkanes always zero.$\endgroup$
– Martin - マーチン♦Sep 10 '14 at 7:46

1

$\begingroup$If you are given a question like this, I think the purpose is to get you to start drawing different structures and realize which ones are equivalent to ones you have already drawn.$\endgroup$
– jerepierreSep 10 '14 at 15:28

5 Answers
5

Determining the number of possible structures for a given range of chemical formulae isn't simple even for saturated hydrocarbons. The number of possible structural isomers rises rapidly with the number of carbons and soon exceeds your ability to enumerate or identify the options by hand. Wikipedia, for example, lists the numbers of isomers and stereoisomers for molecules with up to 120 carbons. But the counts are getting silly even at 10 carbons where there are 75 isomers and 136 stereoisomers.

It has been an interesting research topic in computational chemistry and mathematics. This old paper (pdf), for example, list some formulae for simple hydrocarbons among other simple series. Part of the interest arises because of the relationship to the mathematics of graph theory (it seems that chemistry has inspired some new ideas in this field of mathematics partially because enumerating possible isomers of hydrocarbons is strongly related to drawing certain simple trees which is intuitively obvious if you use the standard chemical convention of drawing just the carbon backbone and ignoring hydrogens).

You can look up the answers on the fascinating mathematics site OEIS (the online encyclopaedia of integer sequences). The sequence for simple hydrocarbons is here.

But the mathematical approach oversimplifies things from the point of view of real-world chemistry. Mathematical trees are idealised abstract objects that ignore real-world chemical constraints like the fact that atoms take up space in three dimensions. This means that some structures that can be drawn cannot exist in the real world because the atoms are too crowded and cannot physically exist without enough strain to cause them to fall apart.

Luckily, computation chemists have also studied this. There is, unfortunately, no obvious shortcut other than trying to create models of the possible structures and testing them to see if they are too strained to exist. The first two isomers that are too crowded are for 16 and 17 carbons and have these structures:

If you have any intuition of the space filling view of these, you should be able to see why they are problematic. A research group at Cambridge University has produced an applet to enumerate the physically possible isomers for a given number of carbons which is available here if your Java settings allow it. The results are discussed in a paper available in the Journal of Chemical Information and Modelling.

$\begingroup$I've ran regression for these wiki data and for long chains number of isomers rises exponentially and it's about 0.001e^x, where x is no of C atoms.$\endgroup$
– MithoronMar 28 '15 at 17:40

The number of alkanes ($\ce{C_nH_{2n+2}}$)
as constitutional isomers (structural isomers) and as steric isomers
is calculated by Polya's theorem (G. Polya and R. C. Read,
Combinatorial Enumeration of Groups, Graphs, and Chemical Compounds, Springer (1987)).
In the process of calculating constitutional isomers,
one 2D structure (graph or constitution) is counted just once.
In the process of calculating steric isomers, one achiral molecule or each chiral molecule
of an enantiomeric pair is counted just once, where achiral molecules and chiral molecules
are not differentiated from each other.

On the other hand,
the number of alkanes ($\ce{C_nH_{2n+2}}$)
as three-dimensional (3D) structural isomers
and as steric isomers is
calculated by Fujita's proligand method
(S. Fujita,
Combinatorial Enumeration of Graphs, Tree-Dimensional Structures, and
Chemical Compounds, Unversity of Kragujevac (2013)).
In the process of calculating 3D structural isomers,
one achiral molecule or one pair of enantiomers is counted just once,
where achiral molecules and chiral molecules (enantiomeric pairs)
are differentiated from each other.

It should be emphasized that graph-theoretical enumerations of chemical compounds
as constitutional isomers (structural isomers) and as steric isomers (based on asymmetry)
should be differentiated from stereochemical enumerations of chemical compounds
as 3D structural isomers and as steric isomers (based on chirality).
Although steric isomers based on asymmetry (graphs governed by permutation groups) and
steric isomers based on chirality (3D structures governed by point groups) give identical
enumeration results, they are conceptually different entities.
This point of view stems from Fujita's stereoisogram approach,
which is described in a recent book
(S. Fujita
Mathematical Stereochemistry, De Gruyter (2015)).

This is a start towards a partial answer, but I tried to count the isomers of methane through nonane, and only by moving methyl groups around, each "branch" is only 1 CH3. I noticed a pattern in the number of 1 branch, 2 branch, 3 branch, and 4 branch isomers. Don't know how accurate these numbers are or how well this pattern holds up.

$\begingroup$The question pertains towards alkanes and the formula that you have specified will be helpful (to a small extent) only for cyclic alkanes. This might as well be a comment than an answer. And Welocme to Chem.SE!$\endgroup$
– Del PateMar 28 '15 at 6:06

$\begingroup$This answer is blatantly wrong. It calculates the double bond equivalents but the example used ($\ce{C4H6}$; i.e. butadiene, cyclobutene or butyne) already has five possible isomers if only a straight chain is assumed. $5 \ne 2$.$\endgroup$
– JanNov 18 '15 at 10:54

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